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sum mXt y = X m<xn,yn is a bounded complex measure on C, with norm 5* ||jt|| • ||j>||, and


mx, x is a bounded measure on C with mass (x \ x). The subspace dom(AO is the set of
all x e E such that the function £>-> |£|2 is mx,x -integrable. For each pair of points

*, >> e dom(AO, we have my>x = mXty,(x\y) =    dmXty&nd(N'X\y)=z \ f(QdmXty(Q.

7.   With the notation of (1 5.12.13) and Problem 6, show that dom(/(AO) is the set of all
x eE such that the function £ *->!/(£) I2 is mx, ^integrable, and that (/(AT)- x\y) =

f /(£) dmx>yQ for all x,ye dom(/(N)). Show that Sp(/(N)) is contained in the closure

of /(Sp(AO). If 9 : C -> C is another universally measurable function, show that
(f°d)(N)-=f(g(N)) (cf. Section 13.9, Problem 21). When N=M/1C) is a simple
continuous normal operator (15.11.3), dom(/(AO) is the set of classes of functions
u e &%(& sucn that/w e ^c(ja), and/(N) is multiplication by the class of/.

8.    Let N be a normal operator and x a point of dom(TV) such that ||jc|| = 1. Put
ax =z (N - x \ x) and jSx =* II N * -x||- Show that, for each e > 0, the open disk Ax with
center a, and radius (j8J -~.|ax|2)1/2 -f e meets Sp(N) (Krylov-Weinstein theorem).

(With the notation of Problem 6, show that j3J— |ax|2 = j |g — ax)2 dmXiXQt and

hence obtain a contradiction of the hypothesis that Ax has measure zero with respect
to mx, x .)

9.    Let jff be a self-adjoint operator on E (bounded or not). For each (bounded or un-
bounded) open interval J <=• R, let P be the orthogonal projector <p*(H). Show that

27T/P, • ;c =    lira      f (((« - ID)/ - F)'1 • jc - ((« + iv)I - H )~l • x) du

u-vO,  u>0 Jj

for all x e £ (" Titchmarsh-Kodaira formula"). (For each j/ e E such that ||j|| g 1,
show that the scalar product of 7 and the difference between 1rriPl • x and the integral
on the right-hand side tends to 0 uniformly for \\y\\ ^ 1 : express this difference in
terms of the measure mx> y (Problem 6), and use the Lebesgue-Fubini theorem and the
dominated convergence theorem.)

10.    Show that, for each closed operator T on E such that dom(T) is dense in E, there
exists a positive self-adjoint operator jR such that dom(JR) = dom(T) and an isometry

V of J«(dom(jR)) onto r(domCO), such that r= VR. (As in the proof of (15.12.8),
define the spaces Fn <= domtT^T) <= dom(r) so that the restriction of T*T to Fn is a
continuous positive self-adjoint operator Hn , for which we may take the square root
Rn . We have || J- xn\\ = \\Rn • xn\\ for all xn e F,, , hence there exists an isometry Vn
of Rtffn) onto T(Fn) such that r|Fn= VnRn. Let R be the positive self-adjoint
operator whose restriction to each Fn is Rn . Show that the Vn are the restrictions of an
isometry V of jR(dom(.R)) onto T(dom(T)), by considering the finite sums of the Fn
and using the facts that Tis closed and domCR) is dense in dom(T). Finally, show that
K-1Tis a closed operator which coincides with Rn on Fd for each n, and hence deduce
that domCR) « dom(T) and V'1 J= R.)

11.    Let T be a closed operator on E such that dom(T) is dense in E and such that
r(dom(T)) c dorn(T*).  Show  that dom(r) = E and that E is continuous. (Ifbspace of E. If T isn E0, to — ToSj;1 on EI, and to —Sn,iT0Snl onEnfor